Understanding the Problem: Triangle Area Ratio with Parallel Lines

The question asks for the ratio of the areas of two triangles, ΔADE and ΔABC, given that DE is a straight line parallel to BC, with points D on AB and E on AC. We are also given the ratio of segments on side AB, AD : DB = 1 : 6.

We know that when a line is drawn parallel to one side of a triangle intersecting the other two sides, it divides the two sides proportionally, and it creates a smaller triangle that is similar to the original triangle.

Analyzing the Triangle Similarity

In ΔABC, the line segment DE is parallel to BC, with D on AB and E on AC.

Consider ΔADE and ΔABC:

• ∠DAE is common to both triangles (∠DAE = ∠BAC).
• Since DE || BC, the corresponding angles are equal: ∠ADE = ∠ABC and ∠AED = ∠ACB.

Because two pairs of corresponding angles are equal, by the Angle-Angle (AA) similarity criterion, ΔADE is similar to ΔABC.

We can write this as: $\Delta ADE \sim \Delta ABC$.

Using Side Ratios in Similar Triangles

For similar triangles, the ratio of corresponding sides is constant. The corresponding sides are AD and AB, AE and AC, and DE and BC.

We are given the ratio AD : DB = 1 : 6. This means that for every 1 unit length of AD, DB has a length of 6 units. Therefore, the total length of AB is AD + DB. If we let AD be $k$ units, then DB is $6k$ units. So, AB = $k + 6k = 7k$ units.

The ratio of the corresponding sides AD and AB is:

$\frac{AD}{AB} = \frac{k}{7k} = \frac{1}{7}$

Since $\Delta ADE \sim \Delta ABC$, the ratio of all corresponding sides is 1/7.

$\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC} = \frac{1}{7}$

Ratio of Areas of Similar Triangles

A fundamental property of similar triangles states that the ratio of their areas is equal to the square of the ratio of their corresponding sides.

Area($\Delta$ADE) / Area($\Delta$ABC) = (Ratio of corresponding sides)$^2$

Using the ratio of sides we found (1/7):

$\frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta ABC)} = \left(\frac{AD}{AB}\right)^2 = \left(\frac{1}{7}\right)^2$

$\frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta ABC)} = \frac{1^2}{7^2} = \frac{1}{49}$

Calculating the Final Ratio

The ratio of the areas of ΔADE and ΔABC is 1/49, which can be written as 1 : 49.

The fact that the area of ΔABC is one unit is additional information that isn't needed to find the ratio of the areas. If we needed the area of ΔADE, we would use this information: Area(ΔADE) / 1 = 1/49, so Area(ΔADE) = 1/49 square units.

Conclusion

Based on the similarity of ΔADE and ΔABC due to DE || BC, and the given ratio AD : DB = 1 : 6 leading to a side ratio AD : AB = 1 : 7, the ratio of their areas is the square of the side ratio.

Ratio of areas = $(1/7)^2 = 1/49$.

Revision Table: Key Concepts for Triangle Area Questions

Additional Information: Extensions and Related Concepts

Understanding triangle similarity and the properties of parallel lines within triangles is crucial for many geometry problems. Here are a few related ideas:

• Converse of BPT: If a line divides two sides of a triangle proportionally, then the line is parallel to the third side.
• Midpoint Theorem: A special case of BPT where the line segment connects the midpoints of two sides. The segment is parallel to the third side and half its length. The area of the small triangle formed is 1/4th the area of the original triangle.
• Altitude Ratio: In similar triangles, the ratio of corresponding altitudes is equal to the ratio of corresponding sides.
• Perimeter Ratio: In similar triangles, the ratio of their perimeters is equal to the ratio of their corresponding sides.

These concepts build upon the fundamental understanding of similarity and proportionality, allowing you to solve a wider range of geometry problems involving triangles and parallel lines.